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The height of a right square pyramid can be similarly obtained, with a substitution of the slant height formula giving: [6] = =. A polyhedron 's surface area is the sum of the areas of its faces. The surface area A {\displaystyle A} of a right square pyramid can be expressed as A = 4 T + S {\displaystyle A=4T+S} , where T {\displaystyle T} and ...
The formula for the volume of a pyramidal square frustum was introduced by the ancient Egyptian mathematics in what is called the Moscow Mathematical Papyrus, written in the 13th dynasty (c. 1850 BC): = (+ +), where a and b are the base and top side lengths, and h is the height.
Given that is the base's area and is the height of a pyramid, the volume of a pyramid is: [29] =. The volume of a pyramid was recorded back in ancient Egypt, where they calculated the volume of a square frustum , suggesting they acquainted the volume of a square pyramid. [ 30 ]
The volume of a tetrahedron can be obtained in many ways. It can be given by using the formula of the pyramid's volume: =. where is the base' area and is the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apices to the opposite faces are inversely proportional to the areas of ...
Because six equilateral square pyramids are removed by truncation, the volume of a truncated octahedron is obtained by subtracting the volume of those six from that of a regular octahedron: [2] = =. The surface area of a truncated octahedron can be obtained by summing all polygonals' area, six squares and eight hexagons.
Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities; List of volume formulas – Quantity of three-dimensional space
For a circular cone with radius r and height h, the base is a circle of area and so the formula for volume becomes [6] V = 1 3 π r 2 h . {\displaystyle V={\frac {1}{3}}\pi r^{2}h.} Slant height
The fourteenth problem of the Moscow Mathematical calculates the volume of a frustum. Problem 14 states that a pyramid has been truncated in such a way that the top area is a square of length 2 units, the bottom a square of length 4 units, and the height 6 units, as shown. The volume is found to be 56 cubic units, which is correct. [1]